Vaart, A. W. van der (1998), Asy

2020-11-15 本文已影响0人陈灿贻

Vaart, A. W. van der (1998), Asymptotic statistics, Cambridge series in statistical and probabilistic mathematics, Cambridge, UK ; New York, NY, USA: Cambridge University Press. Exercise 18.6

It suffice to prove for $[-u,u)$ . For any $\epsilon>0$ , let
$\begin{align*} A& :=\left\{z \in[-u, u]: \exists \delta>0, m\in\mathbb{N}, u_i\in[-u,z], i = 1,\ldots, m, s.t.,\right.\\ & \left. \bigcup_{i = 1}^{m - 1}[u_i, u_{i + 1}) = [-u,z), \forall x, y \in [u_i,u_{i+1}), |x-y|<\delta, |h(x) - h(y)|<\epsilon\right\} \end{align*}$
Let $\mathcal{U}(z):=\{[u_i,u_{i+1}): i = 1,\ldots, m\}$ . Because $h$ is right-continuous at $-u$ , we have $-u\in A$ . Then $A\neq \varnothing$ and $u$ is an upper bound of $A$ . Let $c:=\sup A$ . By the supremum and infimum principle, it suffices to prove $c \in A$ and $c = u$ .

$1^\prime$ Obviously, we have $c\in[-u, u]$ . Because $h$ has left limit in $c$ , there exist $\delta_1>0$ such that for any $x, y \in [c-\delta_1, c)$ , $|h(x) - h(y)|<\epsilon$ . For $\delta_1/2>0$ , by definition of supremum, there exist $\alpha\in A$ such that $\alpha>c - \delta_1/2$ . Without loss of generality, assume $\alpha\leq c$ . Let $\mathcal{U}(c):=\mathcal{U}(x)\cup [\alpha, c)$ . Hence for any $\mathcal{S}\in\mathcal{U}(c)$ , and $x, y\in\mathcal{S}$ , we have $|h(x) -h(y)|<\epsilon$ . It follows $c\in A$ .

$2^\prime$ As for $c = u$ , noting that if $c< u$ , we can find $z:= c + (u-c)3/4$ such that $z>c + (u-c)/2$ and $z\in A$ . This contracts the fact $c=\sup A$ .

Combining $1^\prime$ and $2^\prime$ , we complete the proof.

Vaart, A. W. van der (1998), Asy

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